Problem: Let $R$ be the leftmost region in the first quadrant enclosed by the $x$ -axis and the curve $y=\text{sin}(x)$. $y$ $x$ ${y=\text{sin}(x)}$ ${y=0}$ $y=1}$ $ 0$ $(\pi,0)$ $ R$ A solid is generated by rotating $R$ about the line $y=1$. Which one of the definite integrals gives the volume of the solid? Choose 1 answer: Choose 1 answer: (Choice A) A $\pi \int_0^\pi \left[ 1-\text{sin}^2(x) \right]dx$ (Choice B) B $\pi \int_0^1 \left[ 2\text{sin}(x)-\text{sin}^2(x) \right]dx$ (Choice C) C $\pi \int_0^\pi \left[ 2\text{sin}(x)-\text{sin}^2(x) \right]dx$ (Choice D) D $\pi \int_0^1 \left[ 1-\text{sin}^2(x) \right]dx$
Answer: Let's imagine the solid is made out of many thin slices. Each slice is a cylinder with a hole in the middle, much like a washer. $y$ $x$ ${y=\text{sin}(x)}$ ${y=0}$ $y=1}$ $ 0$ $(\pi,0)$ Let the thickness of each slice be $dx$, let the radius of the washer, as a function of $x$, be $r_1(x)$, and let the radius of the hole, as a function of $x$, be $r_2(x)$. Then, the volume of each slice is $\pi[(r_1(x))^2-(r_2(x))^2]\,dx$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [(r_1(x))^2-(r_2(x))^2]\,dx$ This is called the washer method. What we now need is to figure out the expressions of $r_1(x)$ and $r_2(x)$, and the interval of integration. $r_1(x)$ is equal to the distance between the $x$ -axis and the line $y=1$. So, ${r_1(x)=1}$. $r_2(x)$ is equal to the distance between the curve $y=\text{sin}(x)$ and the line $y=1$. So, ${r_2(x)=1-\text{sin}(x)}$. Now we can find an expression for the area of the washer's base: $\begin{aligned} &\phantom{=} \pi [({r_1(x)})^2-({r_2(x)})^2] \\\\ &= \pi\left[ ({1})^2-\left( {1-\text{sin}(x)} \right)^2 \right] \\\\ &=\pi\left[ 1-\left(1-2\text{sin}(x)+\text{sin}^2(x) \right) \right] \\\\ &=\pi\left[ 2\text{sin}(x)-\text{sin}^2(x) \right] \end{aligned}$ The leftmost endpoint of $R$ is at $x=0$ and the rightmost endpoint is at $x=\pi$. So the interval of integration is $[0,\pi]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_0^\pi \pi\left[ 2\text{sin}(x)-\text{sin}^2(x) \right]dx \\\\ &=\pi \int_0^\pi \left[ 2\text{sin}(x)-\text{sin}^2(x) \right]dx \end{aligned}$